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Find the area enclosed by the parametric equation

  1. Feb 26, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the area of the region enclosed by the parametric equation
    x = t3- 7t
    y = 8t2

    2. Relevant equations
    A = ∫ y(t) x'(t) dt

    3. The attempt at a solution
    I initially began with A = ∫ y(t) x'(t) dt
    And got to ∫24t4-56t2dt and then to 24∫t4dt - 56∫t2dt
    but without a limit/defined area I'm not entirely sure how to proceed.
    At first I assumed the question wanted a general equation, in which case I came up with (24/5)t5 - (56/3)t3, but it doesn't seem to be working.
     
    Last edited: Feb 26, 2017
  2. jcsd
  3. Feb 26, 2017 #2
    It's been a while since I've done this, so someone correct me if I'm wrong. Try factoring ##x(t)=t^3-7t## and ##y(t)=8t^2## and look for the values of ##t## wherein ##x=0## and ##y=0##. Those should be your bounds. Again, it's been a while since I've done calculus so someone feel free to correct me.
     
  4. Feb 26, 2017 #3
    In that case my roots for x would be √7, -√7 and 0, while y would only be 0. What would be my upper and lower bounds?
    I just attempted to make them between 0 and √7 but the answer was wrong as well
     
  5. Feb 26, 2017 #4

    LCKurtz

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    You need to know a bit about what the graph looks like. Apparently it crosses itself, enclosing a loop. If you have a parametric grapher that would be a place to start. Otherwise you can look for a values of ##t## where the graph crosses itself. That would be a values of ##t_1## and ##t_2## where ##(x(t_1),y(t_1)) = (x(t_2),y(t_2))##. Use your equations to figure that out. That will give your limits.
     
  6. Feb 27, 2017 #5

    ehild

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    Eliminate t and find the curve in the (x,y) plane. Yes, it crosses itself at x=0, y=0, which corresponds to t=0. But there is an other crossing point. You get it if you express t as ##t=\pm\sqrt{y/8}## and substitute into the first equation. Because of the plus/minus sign, you get two branches which intersect - where?
     
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